1. Lesson 96 – Expected Value & Variance of Discrete Random Variables HL2 Math - Santowski

2. Opening Exercise: Formulas  If show that

3. Expected Values of Discrete Random Variables  The mean, or expected value, of a discrete random variable is 3

4. Variance of Discrete Random Variables  The variance of a discrete random variable x is  The standard deviation of a discrete random variable x is 4

5. Example #1  You have a “weighted” coin that “favors” the outcome of heads such that p(H) = 0.6  In our experiment, we will have two tosses of this coin and our DRV will be the number of times that heads appears  Determine the probabilities of P(X = 0), P(X = 1) and P(X = 2) and then complete a distribution table and a probability histogram

6. Example #1  You have a “weighted” coin that “favors” the outcome of heads such that p(H) = 0.6 In our experiment, we will have two tosses of this coin and our DRV will be the number of times that heads appears. Our distribution table looks like:  Now we want to calculate some “summary numbers” like  expected value, variance & std deviation x012 P(x)0.160.480.36

7. Summary Measures Calculation Table xp(x)x p(x)x –  Total  (x    p(x) (variance) (x –   (x –   p(x)  x  p(x) (expected value)

8. Summary Measures Calculation Table xp(x)x p(x)x –  Total  (x    p(x) = 0.48 (x –   (x –   p(x)  x  p(x) = 1.2 0 1 2 0.16 0.48 0.36 0 0.48 0.72 0-1.2=-1.2 1-1.2=-0.2 2-1.2=0.8 1.44 0.04 0.64 2.12 0.2304 0.0192 0.2304

9. Example #1 - cont  Now use the results from our calculation to confirm:

10. © 2011 Pearson Education, Inc Summary Measures 1. Expected Value (Mean of probability distribution)  Weighted average of all possible values   = E(x) =  x p(x) 2. Variance  Weighted average of squared deviation about mean   2 = E[(x    (x    p(x) 3. Standard Deviation ●

11. Expected Value  Expected value is an extremely useful concept for good decision-making!

12. Example: the lottery  The Lottery (also known as a tax on people who are bad at math…)  A certain lottery works by picking 6 numbers from 1 to 49. It costs $1.00 to play the lottery, and if you win, you win $2 million after taxes.  If you play the lottery once, what are your expected winnings or losses?

13. Lottery x$p(x).999999928 + 2 million7.2 x 10 --8 Calculate the probability of winning in 1 try: The probability function (note, sums to 1.0): “49 choose 6” Out of 49 numbers, this is the number of distinct combinations of 6.

14. Expected Value x$p(x).999999928 + 2 million7.2 x 10 --8 The probability function Expected Value E(X) = P(win)*$2,000,000 + P(lose)*-$1.00 = 2.0 x 10 6 * 7.2 x 10 -8 +.999999928 (-1) =.144 -.999999928 = -$.86 Negative expected value is never good! You shouldn’t play if you expect to lose money!

15. Expected Value If you play the lottery every week for 10 years, what are your expected winnings or losses?

16. Expected Value If you play the lottery every week for 10 years, what are your expected winnings or losses? 520 x (-.86) = -$447.20

17. Gambling (or how casinos can afford to give so many free drinks…) A roulette wheel has the numbers 1 through 36, as well as 0 and 00. If you bet $1 that an odd number comes up, you win or lose $1 according to whether or not that event occurs. If random variable X denotes your net gain, determine the expected value of X.

18. Gambling (or how casinos can afford to give so many free drinks…) A roulette wheel has the numbers 1 through 36, as well as 0 and 00. If you bet $1 that an odd number comes up, you win or lose $1 according to whether or not that event occurs. If random variable X denotes your net gain, determine the expected value of X. E(X) = 1(18/38) – 1 (20/38) = -$.053 On average, the casino wins (and the player loses) 5 cents per game. The casino rakes in even more if the stakes are higher: E(X) = 10(18/38) – 10 (20/38) = -$.53 If the cost is $10 per game, the casino wins an average of 53 cents per game. If 10,000 games are played in a night, that’s a cool $5300.

19. Practice Problem On the roulette wheel, X=1 with probability 18/38 and X= - 1 with probability 20/38.  We already calculated the mean to be = -$.053. What’s the variance of X?

20. Answer Standard deviation is $.99. Interpretation: On average, you’re either 1 dollar above or 1 dollar below the mean, which is just under zero. Makes sense!

21. Example 2  Our HL Stats & Probability Unit test scores are described by the following probability distribution.  Determine the mean and variance of the scores.  Mr. S, in yet another act of benevolence, decides to scale the scores so my students will not be denied admission to the college of their choice. He decides the actual grades will become:  Grade = 1.5 * Score – 20.  Determine the mean and variance of the grades now. Score4050607080 P(Score).1.2.3.1

22. Example 3  A die is ‘fixed’ so that certain numbers will appear more often. The probability that a 6 appears is twice the probability of a 5 and 3 times the probability of a 4. The probabilities of 3, 2 and 1 are unchanged from a normal die.  The probability distribution table is given below.  Find:  (a) The value of x in the probability distribution and hence complete the probability distribution.  (b) The probability of getting a ‘double’ with two of these dice. Compare with the ‘normal’ probability of getting a double

23. Example 4 Show that, for x = 1,2, …6 is a probability distribution. (a)State P(2 < x < 5) (b)Determine E(X)

24. Example 5  Consider the following gambling game, based on the outcome of the total of 2 dice:  – if the total is a perfect square, you win $4  – if the total is 2, 6, 8 or 10, you win $1  – otherwise, you lose $2.  (a) Find the expected value of this game.  (b) Determine if it is a fair game.

25. Example 6  Find the missing profit (or loss) so that the following probability table has an expected value of 0.

26. Example 7  For the following probability distribution calculate:  E(X)  E(2X)  E(X + 2)  E(X 2 )  E(X 2 ) – [E(X)] 2.

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